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3Sum

Difficulty: Medium

Problem Description

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0. Notice that the solution set must not contain duplicate triplets.

Key Insights

  • The problem requires finding unique triplets in an array that sum to zero.
  • Sorting the array helps in efficiently finding pairs that complement a given number.
  • Using a two-pointer approach reduces the time complexity when searching for pairs.

Space and Time Complexity

Time Complexity: O(n^2), where n is the number of elements in the input array. This is due to the nested iteration over the array after sorting. Space Complexity: O(1) for the pointers; O(n) for the output list storing the triplets.

Solution

The algorithm begins by sorting the input array. Then, for each element, it uses a two-pointer technique to find pairs that sum up to the negative of the current element. This approach efficiently eliminates duplicates by checking adjacent elements after sorting.

Code Solutions

def threeSum(nums):
    nums.sort()  # Sort the array to facilitate the two-pointer approach
    result = []
    n = len(nums)

    for i in range(n):
        if i > 0 and nums[i] == nums[i - 1]:  # Skip duplicates
            continue

        left, right = i + 1, n - 1  # Initialize two pointers

        while left < right:
            total = nums[i] + nums[left] + nums[right]
            if total < 0:
                left += 1  # Move left pointer to the right to increase sum
            elif total > 0:
                right -= 1  # Move right pointer to the left to decrease sum
            else:
                result.append([nums[i], nums[left], nums[right]])  # Found a triplet
                while left < right and nums[left] == nums[left + 1]:  # Skip duplicates
                    left += 1
                while left < right and nums[right] == nums[right - 1]:  # Skip duplicates
                    right -= 1
                left += 1
                right -= 1

    return result
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